Centroids of triangles inscribed/circumscribed to 2 circles.

freemind

Centroids of triangles inscribed/circumscribed to 2 circles.

**Posted:** Mon Mar 03, 2008 6:04 pm

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Amir.S

the nine-point circles of $A_iB_iC_i$ are tangent to incircle with a constant Radius hence the locus of $H$ (homothecy point of Circumcircle and ninpoint circle) is a cirlce hence the locus of centroids is a cricle.

**Posted:** Tue Mar 04, 2008 9:20 pm

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Inequalities Master · Yang-Mills Theory Offline Joined: 16 Feb 2008 Posts: 572

#3

Can you explain with more details your solution,please?I am weak at geometry and I do not know many things.Thanks Sad

**Posted:** Sat Mar 29, 2008 4:02 pm

Amir.S

#4

**Posted:** Sun Mar 30, 2008 12:28 am

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Say: Tell me if it is from Allah; then you disbelieve in it, who is in greater error than he who is in a prolonged opposition?

Inequalities Master · Yang-Mills Theory Offline Joined: 16 Feb 2008 Posts: 572

#5

Sorry,but I think I need more details.Sorry for the late answer. Wallbash

**Posted:** Thu Apr 10, 2008 2:57 pm

freemind

#6

Here's my solution from the TST:

Apply a symmetry to the triangle wrt to $OI$ . Clearly, the new triangle, if different from the original will also be circumscribed/inscribed to $\Gamma(I,r)$ and $\Gamma(O,R)$ respectively. Hence for the problem to be true, we need to show that all centroids lie on a circle of center lying on $OI$ . Assume we have found this center $M$ . Moreover assume it lies on $(OI)$ . Let $IM/IO = \lambda$ . For $MG$ to be a constant value (not depending on the chosen triangle), we are looking to express $MG$ as a function only of $R$ and $r$ (not depending on other elements of the triangle). Since $R$ and $r$ remain constant, the solution will be done. In fact we are looking for a constant $\lambda$ so that $MG = f(R,r)$ .

So here we go:
From Stewart Relation $GM^2\cdot OI = GI^2\cdot MO + GO^2\cdot IM - IM\cdot MO\cdot IO$ , or $GM^2 = (1 - \lambda)GI^2 + \lambda GO^2 - OI^2\lambda(1 - \lambda)$ .
Well, $OI^2\lambda(1 - \lambda) = g(R,r)$ , hence we need only look at $E = (1 - \lambda)GI^2 + \lambda GO^2$ . Use Leibniz to obtain
$3E = (1 - \lambda)\left [IA^2 + IB^2 + IC^2 - \frac13(a^2 + b^2 + c^2)\right] + \lambda\left[3R^2 - \frac13(a^2 + b^2 + c^2)\...$ .

Hence we are looking for a $\lambda$ so that $E' = 3(1 - \lambda)(IA^2 + IB^2 + IC^2) - (a^2 + b^2 + c^2)$ to equal $g(R,r)$ . Let $3(1 - \lambda) = \alpha$ . Note that $IA^2 + IB^2 + IC^2 = 3r^2 + (p - a)^2 + (p - b)^2 + (p - c)^2$ , so
$E' = 3\alpha r^2 + \alpha[(p - a)^2 + (p - b)^2 + (p - c)^2] - (a^2 + b^2 + c^2)$ . Finally, we must find an $\alpha$ so that $F = \alpha[(p - a)^2 + (p - b)^2 + (p - c)^2] - (a^2 + b^2 + c^2) = h(R,r)$ .
Using $a + b + c = 2p$ and $ab + bc + ca = r^2 + p^2 + 4Rr$ , we get
$F = - \alpha p^2 + (\alpha - 1)(\sum a^2) = - \alpha p^2 + (\alpha - 1)[4p^2 - 2(r^2 + p^2 + 4Rr)]$ . Further $F = Q(R,r) + (\alpha - 2)p^2$ .
So taking $\alpha = 2$ , works. That means, $\lambda = \frac13$ . Hence, all centroids lie on a circle with center $M\in(OI)$ , so that $IO = 3IM$ . Moreover, tracking back the 'ignored' functions of $R$ and $r$ , we easily get $GM^2 = \frac {(R - 2r)^2}9 = \frac49\left(\frac R2 - r\right)^2$ .
Actually, as I now realize, using $\omega O = 3\omega G$ , the above is a proof for $I\omega = \frac R2 - r$ , from which follows Feuerbach's Theorem: the Nine-Point Circle is tangent to the incircle (because radius of 9-point circle is $\frac R2$ ).

**Posted:** Fri Apr 11, 2008 7:11 pm

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